How do you find the vertex, focus and directrix of #x+y^2=0#?

1 Answer
Aug 3, 2018

Vertex is at #(0,0)#, focus is at #(-0.25,0)# and
directrix is
#x= 0.25#

Explanation:

# x+y^2=0 or y^2 = -x or (y-0)^2 = -4 *1/4 (x-0)#

The equation of horizontal parabola opening left is

#(y-k)^2 = -4 p(x-h) ; h=0 ,k=0 ; p= 1/4=0.25#

Therefore, vertex is at #(0,0)# The distance between focus and

vertex is #p=0.25 :.# Focus is at #(-0.25,0)# . Vertex is at midway

between focus and directrix. Therefore , directrix is #x= 0.25#

graph{x+y^2=0 [-12.66, 12.65, -6.33, 6.33]} [Ans]